Finding all values $p$ for which $int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$ converges












1












$begingroup$


I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
$$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
converges. So far I've came up with this:
$$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
\
text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
\
lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
\
int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
$$
So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?






calculus integration convergence improper-integrals







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    1












    $begingroup$


    I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
    $$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
    converges. So far I've came up with this:
    $$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
    \
    text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
    \
    lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
    \
    int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
    $$
    So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?






    calculus integration convergence improper-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
      $$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
      converges. So far I've came up with this:
      $$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
      \
      text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
      \
      lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
      \
      int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
      $$
      So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?






      calculus integration convergence improper-integrals







      share|cite|improve this question











      $endgroup$




      I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
      $$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
      converges. So far I've came up with this:
      $$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
      \
      text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
      \
      lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
      \
      int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
      $$
      So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?






      calculus integration convergence improper-integrals




      calculus integration convergence improper-integrals






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      edited Nov 30 '18 at 5:40









      Olivier Oloa

      108k17177294




      108k17177294










      asked Nov 30 '18 at 0:20









      twkmztwkmz

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          $begingroup$

          Hint. One may use that (with an integration by parts)
          $$
          int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
          $$           then one may observe that, as $x to infty$,
          $$
          frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
          $$






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            1 Answer
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            1 Answer
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            active

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            3












            $begingroup$

            Hint. One may use that (with an integration by parts)
            $$
            int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
            $$             then one may observe that, as $x to infty$,
            $$
            frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
            $$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Hint. One may use that (with an integration by parts)
              $$
              int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
              $$               then one may observe that, as $x to infty$,
              $$
              frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
              $$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Hint. One may use that (with an integration by parts)
                $$
                int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
                $$                 then one may observe that, as $x to infty$,
                $$
                frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
                $$






                share|cite|improve this answer









                $endgroup$



                Hint. One may use that (with an integration by parts)
                $$
                int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
                $$                 then one may observe that, as $x to infty$,
                $$
                frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 '18 at 0:26









                Olivier OloaOlivier Oloa

                108k17177294




                108k17177294






























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